(1– (C+D)), provided (C+D) < 1. where K1 Assume that P and Q are two positive distinct real roots of the quadratic equation 2- (P+ Q)t+ PQ=0. (27) Thus, we deduce that (P+ Q)² > 4PQ. (28) Substituting (25) and (26) into (28), we get the condition (20). Thus, the proof is now completed.O
(1– (C+D)), provided (C+D) < 1. where K1 Assume that P and Q are two positive distinct real roots of the quadratic equation 2- (P+ Q)t+ PQ=0. (27) Thus, we deduce that (P+ Q)² > 4PQ. (28) Substituting (25) and (26) into (28), we get the condition (20). Thus, the proof is now completed.O
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Topic Video
Question
Explain the determine yellow and the inf is here
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxp=k– eXp–1]
Xn+1 =
Axn+ Bxn-k+ Cxp–1+Dxp-
n= 0,1,2, ....
(1)
where the coefficients A, B, C, D, b, d, e E (0, 0), while
k, 1 and o are positive integers. The initial conditions
i,.……, X_1,..., X_ky •……, X_1, Xo are arbitrary positive real
numbers such that k < 1< 0. Note that the special cases
of Eq.(1) have been studied in [1] when B= C = D= 0,
and k = 0,1= 1, b is replaced by – b and in [27] when
B=C= D=0, and k= 0, b is replaced by
[33] when B = C = D = 0, 1= 0 and in [32] when
A= C=D=0, 1=0, b is replaced by – b.
b and in
|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Fab26bc06-b34a-4ec6-9aaf-5565ced1cb84%2Fnhr6yt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxp=k– eXp–1]
Xn+1 =
Axn+ Bxn-k+ Cxp–1+Dxp-
n= 0,1,2, ....
(1)
where the coefficients A, B, C, D, b, d, e E (0, 0), while
k, 1 and o are positive integers. The initial conditions
i,.……, X_1,..., X_ky •……, X_1, Xo are arbitrary positive real
numbers such that k < 1< 0. Note that the special cases
of Eq.(1) have been studied in [1] when B= C = D= 0,
and k = 0,1= 1, b is replaced by – b and in [27] when
B=C= D=0, and k= 0, b is replaced by
[33] when B = C = D = 0, 1= 0 and in [32] when
A= C=D=0, 1=0, b is replaced by – b.
b and in
|
![Theorem 9.If k is even and 1, 0 are odd positive integers,
then Eq.(1) has prime period two solution if the condition
t
(1–(C+D)) (3e– d) < (e+ d) (A+ B),
(20)
is
valid,
provided
(C+D)
1
and
e (1– (C+ D)) – d (A+B) > 0.
a
Proof.If k is even and 1, o are odd positive integers, then
Xn = Xn-k and xn+1 = Xn-1= Xn-o• It follows from Eq.(1)
that
bQ
P= (A+B) Q+(C+D)P –
(21)
(еР — dQ)"
and
V
БР
Q= (A+B) P+(C+D) Q –
(22)
(eQ– dP)'
Consequently, we get
e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+ D) P
- (C+D) dPQ– bQ,
(23)
and
e Q – dPQ = e (A+B) PQ– d (A+B) +e(C+D) Q
- (C+D) dPQ– bP.
i
%3D
(24)
By subtracting (24) from (23), we get
P+Q=
(25)
[e (1– (C+D))– d (A+B)]’
where e (1-(C+ D)) – d (A+B) > 0. By adding (23)
and (24), we obtain
t
e b (1 – (C+D))
(e+d) [K1+(A+B)][e K1 – d (A+B)]² '
PQ=
(26)
(1–(C+D)), provided (C+D) < 1.
where K1 =
Assume that P and Q are two positive distinct real roots
of the quadratic equation
2- (P+Q)t+ PQ=0.
(27)
Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Fab26bc06-b34a-4ec6-9aaf-5565ced1cb84%2Fb95qr4m_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 9.If k is even and 1, 0 are odd positive integers,
then Eq.(1) has prime period two solution if the condition
t
(1–(C+D)) (3e– d) < (e+ d) (A+ B),
(20)
is
valid,
provided
(C+D)
1
and
e (1– (C+ D)) – d (A+B) > 0.
a
Proof.If k is even and 1, o are odd positive integers, then
Xn = Xn-k and xn+1 = Xn-1= Xn-o• It follows from Eq.(1)
that
bQ
P= (A+B) Q+(C+D)P –
(21)
(еР — dQ)"
and
V
БР
Q= (A+B) P+(C+D) Q –
(22)
(eQ– dP)'
Consequently, we get
e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+ D) P
- (C+D) dPQ– bQ,
(23)
and
e Q – dPQ = e (A+B) PQ– d (A+B) +e(C+D) Q
- (C+D) dPQ– bP.
i
%3D
(24)
By subtracting (24) from (23), we get
P+Q=
(25)
[e (1– (C+D))– d (A+B)]’
where e (1-(C+ D)) – d (A+B) > 0. By adding (23)
and (24), we obtain
t
e b (1 – (C+D))
(e+d) [K1+(A+B)][e K1 – d (A+B)]² '
PQ=
(26)
(1–(C+D)), provided (C+D) < 1.
where K1 =
Assume that P and Q are two positive distinct real roots
of the quadratic equation
2- (P+Q)t+ PQ=0.
(27)
Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O
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